If $f(z)=\sum c_nz^n$ is an entire function of finite genus $\mu$ then $\lim_{n\to\infty}c_n(n!)^{1/(\mu+1)}=0$ If $f(z)=\sum c_nz^n$ is an entire function of finite genus $\mu$ then prove that $$\lim_{n\to\infty}c_n(n!)^{1/(\mu+1)}=0.$$
I know that if $f$ is an entire function of finite genus $\mu$, then $f$ is of finite order $\lambda\leq\mu+1$. In particular, I have $|f(z)|\lt \exp(|z|^{\beta})$ holds for all large enough $|z|$, where $\beta=\mu+1$. Now, using a previous estimate, I get $$|c_n|\leq \Bigl(\frac{e\beta}{n}\Bigl)^{n/\beta}$$ for all large enough $n$. Thus, 
$$\begin{align*}
|c_n|(n!)^{1/\beta}
&\leq \Bigl(\frac{e\beta}{n}\Bigl)^{n/\beta}(n!)^{1/\beta}\\
&\sim \Bigl(\frac{e\beta}{n}\Bigl)^{n/\beta}\Bigl(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\Bigl)^{1/\beta}\\
&=\beta^{n/\beta}(2 \pi n)^{1/2\beta}.
\end{align*}$$
However, this doesn't help much. I don't think $\beta^{n/\beta}(2 \pi n)^{1/2\beta}$ converges to $0$.
Now, if I just use the Cauchy's estimate, I have 
\begin{align*}
|c_n|(n!)^{1/\beta}&\leq \frac{\exp(n^{\beta})}{n^n} (n!)^{1/\beta}\\
&\sim \frac{\exp(n^{\beta})}{n^n}\Bigl(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\Bigl)^{1/\beta}\\
&=(2 \pi)^{1/2\beta}\exp(n^{\beta}-n/\beta)n^{n/\beta+1/2\beta-n}.
\end{align*}
I don't have much experience in computing estimates. Can someone help me out to complete any of the arguments?

This is Exercise $2.1$ from Chapter ${\rm XI}$ of Conway's functions of one complex variable. The hint given is to use the Cauchy's estimates. I found this AoPS post using approach0, however, they mistook the genus for the order. Also, the definition of the "genus"(order) mentioned in the above post comes only after this exercise in Conway(Exercise $2.5$). 
Here's the definition of the order Conway uses:
Definition. An entire function $f$ is of finite order if there is a positive constant $a$ and an $r_0\gt0$ such that $|f(z)|\lt \exp(|z|^a)$ for $|z|\gt r_0$. If $f$ is of finite order, then $$\lambda=\inf\{a\mid |f(z)|\lt \exp(|z|^a)\,\text{for $|z|$ sufficiently large}\}$$ is called the order of $f$. Equivalently, $$\lambda=\limsup_{r\to\infty}\frac{\log\log M(r)}{\log r},$$ where $M(r)=\max\{|f(z)|\mid|z|=r\}$.
Of course, all three are equivalent, but I would like to see a proof using the hint and the above two definitions.
 A: Partial answer. Suppose that the order of $f(z)$ is $\lambda$. As you already noted, $\lambda  \le \mu  + 1$. Let us consider the case that $\lambda <\mu+1$. Then
$$
\frac{1}{\lambda } - \frac{1}{{\mu  + 1}} = \varepsilon,
$$
with a suitable positive $\varepsilon$. Since
$$
\mathop {\lim \inf }\limits_{n \to  + \infty } \frac{{ - \log \left| {c_n } \right|}}{{n\log n}} = \frac{1}{\lambda },
$$
it holds that
$$
\frac{{ - \log \left| {c_n } \right|}}{{n\log n}} \geq \frac{1}{\lambda } -\frac{\varepsilon }{2},
$$
for all $n\geq n_0$ with a suiable $n_0\geq 0$ that depends on $\varepsilon$. Hence, for all $n\geq n_0\geq 0$,
\begin{align*}
c_n (n!)^{1/(\mu  + 1)} & = \exp \left( { - \left[ {\frac{{ - \log \left| {c_n } \right|}}{{n\log n}} - \frac{1}{{\mu  + 1}}\frac{{\log n!}}{{n\log n}}} \right]n\log n} \right)
\\ &
 \le \exp \left( { - \left[ {\frac{1}{\lambda } - \frac{\varepsilon }{2} - \frac{1}{{\mu  + 1}}\frac{{\log n!}}{{n\log n}}} \right]n\log n} \right) \\ & = \exp \left( { - \left[ {\frac{\varepsilon }{2} + \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right)} \right]n\log n} \right) .
\end{align*}
Taking the limit of both sides, yields the claim.
